Projection of point and line
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NOTATIONS
FOLLOWING NOTATIONS SHOULD BE FOLLOWED WHILE NAMEING
DIFFERENT VIEWS IN ORTHOGRAPHIC PROJECTIONS.
OBJECT
POINT A
LINE AB
IT’S TOP VIEW
a
ab
IT’S FRONT VIEW
a’
a’ b’
IT’S SIDE VIEW
a”
a” b”
SAME SYSTEM OF NOTATIONS SHOULD BE FOLLOWED
INCASE NUMBERS, LIKE 1, 2, 3 – ARE USED.
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VP
2nd
Quad.
1ST Quad.
Y
Observer
X Y
HP
X
3rd Quad.
4th Quad.
THIS QUADRANT PATTERN,
IF OBSERVED ALONG X-Y LINE ( IN RED ARROW DIRECTION)
WILL EXACTLY APPEAR AS SHOWN ON RIGHT SIDE AND HENCE,
IT IS FURTHER USED TO UNDERSTAND ILLUSTRATION PROPERLLY.
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PROJECTIONS OF A POINT IN FIRST QUADRANT.
POINT A ABOVE HP
& IN VP
POINT A ABOVE HP
& INFRONT OF VP
For Tv
For Tv
PICTORIAL
PRESENTATION
a’
a’
PICTORIAL
PRESENTATION
A
A
For Tv
Y
Y
X
POINT A IN HP
& INFRONT OF VP
a’
a
a
X
X
a
Y
A
ORTHOGRAPHIC PRESENTATIONS
OF ALL ABOVE CASES.
Fv above xy,
Tv below xy.
Fv above xy,
Tv on xy.
VP
Fv on xy,
Tv below xy.
VP
a’
X
VP
a’
Y
X
a
Y
a’
X
a
HP
a
HP
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HP
Y
POINT A IN
Point A is
ND
2 QUADRANT
Placed In
A
different
quadrants
and it’s Fv & Tv
are brought in
same plane for
Observer to see
HP
clearly.
Fv is visible as
it is a view on
VP. But as Tv is
is a view on Hp,
it is rotated
downward 900,
In clockwise
direction.The
In front part of
Hp comes below
xy line and the
HP
part behind Vp
comes above.
Observe and
note the
process.
A
POINT A IN
RD
3
QUADRANT
POINT A IN
1ST QUADRANT
VP
a’
VP
a’
A
a
HP
OBSERVER
OBSERVER
a
a
HP
OBSERVER
OBSERVER
a
a’
a’
VP
VP
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A
POINT A IN
4TH QUADRANT
PROJECTIONS OF STRAIGHT LINES.
INFORMATION REGARDING A LINE means
IT’S LENGTH,
POSITION OF IT’S ENDS WITH HP & VP
IT’S INCLINATIONS WITH HP & VP WILL BE GIVEN.
AIM:- TO DRAW IT’S PROJECTIONS - MEANS FV & TV.
SIMPLE CASES OF THE LINE
1.
A VERTICAL LINE ( LINE PERPENDICULAR TO HP & // TO VP)
2.
LINE PARALLEL TO BOTH HP & VP.
3.
LINE INCLINED TO HP & PARALLEL TO VP.
4.
LINE INCLINED TO VP & PARALLEL TO HP.
5.
LINE INCLINED TO BOTH HP & VP.
STUDY ILLUSTRATIONS GIVEN ON NEXT PAGE
SHOWING CLEARLY THE NATURE OF FV & TV
OF LINES LISTED ABOVE AND NOTE RESULTS.
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For Tv
Fv is a vertical line
Showing True
Length &
Tv is a point.
a’
A
1.
FV
A Line
perpendicular
to Hp
&
// to Vp
b’
Y
Orthographic Pattern
V.P.
a’
Fv
b’
X
Y
B
TV a b
Tv a b
X
H.P.
Orthographic Pattern
V.P.
For Tv
Fv & Tv both are
// to xy
&
both show T. L.
2.
b’
A Line
// to Hp
&
// to Vp
B
a’
A
a’
Fv
b’
X
Y
b
Y
a
Tv
X
a
H.P.
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b
b’
3.
A Line inclined to Hp
and
parallel to Vp
V.P.
Fv inclined to xy
Tv parallel to xy.
b’
B
a’

a’

Y
X
Y

A
a
b
T.V.
b
X
a
H.P.
Orthographic Projections
4.
A Line inclined to Vp
and
parallel to Hp
V.P.
Tv inclined to xy
Fv parallel to xy.
a’
b’
Fv
b’
a’
A
Ø
B
X
Y
a
Ø
a
Ø
Tv
b
b
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H.P.
For Tv
For Tv
5.
b’
A Line inclined to both
Hp and Vp
b’
(Pictorial presentation)
B
B


Y
A
X
a’
On removal of object
i.e. Line AB
Fv as a image on Vp.
Tv as a image on Hp,
a’

a
A
X
T.V.
Y
b

a
T.V.
b
V.P.
b’
FV
a’

X
Orthographic Projections
Fv is seen on Vp clearly.
To see Tv clearly, HP is
rotated 900 downwards,
Hence it comes below xy.
Y
a

TV
H.P.
b
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Note These Facts:Both Fv & Tv are inclined to xy.
(No view is parallel to xy)
Both Fv & Tv are reduced
lengths.
(No view shows True Length)
Note the procedure
When Fv & Tv known,
How to find True Length.
(Views are rotated to determine
True Length & it’s inclinations
with Hp & Vp).
Orthographic Projections
Means Fv & Tv of Line AB
are shown below,
with their apparent Inclinations
&
V.P.
V.P.
V.P.
b’
b’
FV
a’

a’
Y


TL
a’ 
X
Y
a
TV
H.P.
b’ b1’
b 1’
FV
X
a
Note the procedure
When True Length is known,
How to locate Fv & Tv.
(Component a-1 of TL is drawn
which is further rotated
to determine Fv)

TV
b2

1’
X
Y
a
1
Ø

TV
b
Here TV (ab) is not // to XY line
Hence it’s corresponding FV
a’ b’ is not showing
True Length &
True Inclination with Hp.
H.P.
b
In this sketch, TV is rotated
and made // to XY line.
Hence it’s corresponding
FV a’ b1’ Is showing
True Length
&
True engineering108.com
Inclination with Hp.
H.P.
b
b1
Here a -1 is component
of TL ab1 gives length of Fv.
Hence it is brought Up to
Locus of a’ and further rotated
to get point b’. a’ b’ will be Fv.
Similarly drawing component
of other TL(a’ b1‘) Tv can be drawn.
The most important diagram showing graphical relations
among all important parameters of this topic.
Study and memorize it as a CIRCUIT DIAGRAM
And use in solving various problems.
1) True Length ( TL) – a’ b1’ & a b
2) Angle of TL with Hp -

3) Angle of TL with Vp –
Ø
4) Angle of FV with xy –
5) Angle of TV with xy –
V.P.
Distance between
End Projectors.

6) LTV (length of FV) – Component (a-1)
b1’
b’

Important
TEN parameters
to be remembered
with Notations
used here onward
7) LFV (length of TV) – Component (a’-1’)
8) Position of A- Distances of a & a’ from xy
9) Position of B- Distances of b & b’ from xy
a’

10) Distance between End Projectors

1’
LTV
X
Y
LFV
a
Ø
1
NOTE this
& 
Construct with a’
Ø&
Construct with a
b’ & b1’ on same locus.

b & b1 on same locus.
H.P.
b
b1
Also Remember
True Length is never rotated. It’s horizontal component is
drawn & it is further rotated to locate view.
Views are always rotated, made horizontal & further
extended to locate TL,  & Ø
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GROUP (A)
PROBLEM
GENERAL CASES OF THE LINE INCLINED TO BOTH HP & VP
( based on 10 parameters).
Line AB is 75 mm long and it is 300 &
400 Inclined to Hp & Vp respectively.
End A is 12mm above Hp and 10 mm
in front of Vp.
Draw projections. Line is in 1st quadrant.
b’
FV
SOLUTION STEPS:
1) Draw xy line and one projector.
2) Locate a’ 12mm above xy line
& a 10mm below xy line.
3) Take 300 angle from a’ & 400 from
a and mark TL I.e. 75mm on both
lines. Name those points b1’ and b1
respectively.
4) Join both points with a’ and a resp.
5) Draw horizontal lines (Locus) from
both points.
6) Draw horizontal component of TL
a b1 from point b1 and name it 1.
( the length a-1 gives length of Fv as
we have seen already.)
7) Extend it up to locus of a’ and rotating
a’ as center locate b’ as shown.
Join a’ b’ as Fv.
8) From b’ drop a projector down ward
& get point b. Join a & b I.e. Tv.
b’1
TL

a’
X
Y
a
LFV
Ø
1
TV
TL
b
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b1
PROBLEM :
Line AB 75mm long makes 450 inclination with Vp while it’s Fv makes 550.
End A is 10 mm above Hp and 15 mm in front of Vp.If line is in 1st quadrant
draw it’s projections and find it’s inclination with Hp.
b’
b’1 LOCUS OF
Solution Steps:1.Draw x-y line.
2.Draw one projector for a’ & a
3.Locate a’ 10mm above x-y &
Tv a 15 mm below xy.
4.Draw a line 450 inclined to xy
from point a and cut TL 75 mm
on it and name that point b1
Draw locus from point b1
5.Take 550 angle from a’ for Fv
above xy line.
6.Draw a vertical line from b1
up to locus of a and name it 1.
It is horizontal component of
TL & is LFV.
7.Continue it to locus of a’ and
rotate upward up to the line
of Fv and name it b’.This a’ b’
line is Fv.
8. Drop a projector from b’ on
locus from point b1 and
name intersecting point b.
Line a b is Tv of line AB.
9.Draw locus from b’ and from
a’ with TL distance cut point b1‘
10.Join a’ b1’ as TL and measure
it’s angle at a’.
It will be true angle of line with HP.
550
a’
y
X
LFV
a
1
LOCUS OF b
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b
b1
PROBLEM - Fv of line AB is 500 inclined to xy and measures 55 mm long while it’s
Tv is 600 inclined to xy line.
If end A is 10 mm above Hp and 15 mm in front of
Vp, draw it’s projections,find TL, inclinations of line with Hp & Vp.
b’
SOLUTION STEPS:
1.Draw xy line and one projector.
2.Locate a’ 10 mm above xy and
X
a 15 mm below xy line.
3.Draw locus from these points.
4.Draw Fv 500 to xy from a’ and
mark b’ Cutting 55mm on it.
5.Similarly draw Tv 600 to xy
from a & drawing projector from b’
Locate point b and join a b.
6.Then rotating views as shown,
locate True Lengths ab1 & a’b1’
and their angles with Hp and Vp.
a’

b’1
500
y
a

600
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b
b1
PROBLEM :Line AB is 75 mm long .It’s Fv and Tv measure 50 mm & 60 mm long respectively.
End A is 10 mm above Hp and 15 mm in front of Vp. Draw projections of line AB
if end B is in first quadrant. Find angle with Hp and Vp.
b’
SOLUTION STEPS:
1.Draw xy line and one projector.
2.Locate a’ 10 mm above xy and
a 15 mm below xy line.
3.Draw locus from these points.

4.Cut 60mm distance on locus of a’
a’
& mark 1’ on it as it is LTV.
X
5.Similarly Similarly cut 50mm on
locus of a and mark point 1 as it is LFV.
a
6.From 1’ draw a vertical line upward

and from a’ taking TL ( 75mm ) in
compass, mark b’1 point on it.
Join a’ b’1 points.
7. Draw locus from b’1
8. With same steps below get b1 point
and draw also locus from it.
9. Now rotating one of the components
I.e. a-1 locate b’ and join a’ with it
to get Fv.
 & 
10. Locate tv similarly and measure
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Angles
b’1
1’
LTV
Y
LFV
b
1
b1
PROBLEM :-
T.V. of a 75 mm long Line CD, measures 50 mm.
End C is in Hp and 50 mm in front of Vp.
End D is 15 mm in front of Vp and it is above Hp.
d’
Draw projections of CD and find angles with Hp and Vp.
SOLUTION STEPS:
1.Draw xy line and one projector.
2.Locate c’ on xy and
c 50mm below xy line.
3.Draw locus from these points.
4.Draw locus of d 15 mm below xy c’
5.Cut 50mm & 75 mm distances X
on
locus of d from c and mark points
d & d1 as these are Tv and line CD
lengths resp.& join both with c.
6.From d1 draw a vertical line upward
up to xy I.e. up to locus of c’ and
draw an arc as shown.
 &  from d to
7 Then draw one projector
meet this arc in d’ point & join c’ d’
8. Draw locus of d’ and cut 75 mm c
on it from c’ as TL
9.Measure Angles
d’1
LOCUS OF d’ & d’1

Y
d

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d1
LOCUS OF d & d1
GROUP (B)
PROBLEMS INVOLVING TRACES OF THE LINE.
TRACES OF THE LINE:THESE ARE THE POINTS OF INTERSECTIONS OF A LINE ( OR IT’S EXTENSION )
WITH RESPECTIVE REFFERENCE PLANES.
A LINE ITSELF OR IT’S EXTENSION, WHERE EVER TOUCHES H.P.,
THAT POINT IS CALLED TRACE OF THE LINE ON H.P.( IT IS CALLED H.T.)
SIMILARLY, A LINE ITSELF OR IT’S EXTENSION, WHERE EVER TOUCHES V.P.,
THAT POINT IS CALLED TRACE OF THE LINE ON V.P.( IT IS CALLED V.T.)
V.T.:-
It is a point on Vp.
Hence it is called Fv of a point in Vp.
Hence it’s Tv comes on XY line.( Here onward named as v
H.T.:-
)
It is a point on Hp.
Hence it is called Tv of a point in Hp.
Hence it’s Fv comes on XY line.( Here onward named as ’h’
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)
b’
STEPS TO LOCATE HT.
(WHEN PROJECTIONS ARE GIVEN.)
1.
2.
3.
4.
Begin with FV. Extend FV up to XY line.
Name this point h’
( as it is a Fv of a point in Hp)
Draw one projector from h’.
Now extend Tv to meet this projector.
This point is HT
STEPS TO LOCATE VT.
(WHEN PROJECTIONS ARE GIVEN.)
1.
Begin with TV. Extend TV up to XY line.
2.
Name this point v
( as it is a Tv of a point in Vp)
Draw one projector from v.
Now extend Fv to meet this projector.
This point is VT
3.
4.
a’
v
h’
VT’
HT
x
y
a
Observe & note :1. Points h’ & v always on x-y line.
2. VT’ & v always on one projector.
3. HT & h’ always on one projector.
4. FV - h’- VT’ always co-linear.
5. TV - v - HT always co-linear.
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b
PROBLEM :- Fv of line AB makes 450 angle
with XY line and measures 60 mm.
Line’s Tv makes 300 with XY line. End A is 15
mm above Hp and it’s VT is 10 mm
below Hp. Draw projections of line
AB,determine inclinations with Hp & Vp and
locate HT, VT.
SOLUTION STEPS:a’
Draw xy line, one projector and
15
locate fv a’ 15 mm above xy.
v h’
x
Take 450 angle from a’ and
300
marking 60 mm on it locate point 10
b’.
Draw locus of VT, 10 mm below xy
VT’
& extending Fv to this locus locate VT.
a
as fv-h’-vt’ lie on one st.line.
Draw projector from vt, locate v on xy.
From v take 300 angle downward as
Tv and it’s inclination can begin with v.
Draw projector from b’ and locate b I.e.Tv point.
Now rotating views as usual TL and
it’s inclinations can be found.
Name extension of Fv, touching xy as h’
and below it, on extension of Tv, locate HT.
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b’

b’1
450
y

b
b1
PROBLEM
One end of line AB is 10mm above Hp and
other end is 100 mm in-front of Vp.
It’s Fv is 450 inclined to xy while it’s HT & VT
are 45mm and 30 mm below xy respectively.
Draw projections and find TL with it’s
inclinations with Hp & VP.
b’
a’
10
X
v

b’1
LOCUS OF b’ & b’1
450
h’
Y
30
SOLUTION STEPS:45
VT’
Draw xy line, one projector and
HT
locate a’ 10 mm above xy.
Draw locus 100 mm below xy for points b & b1
Draw loci for VT and HT, 30 mm & 45 mm
a

below xy respectively.
Take 450 angle from a’ and extend that line backward
to locate h’ and VT, & Locate v on xy above VT.
Locate HT below h’ as shown.
Then join v – HT – and extend to get top view end b.
Draw projector upward and locate b’ Make a b & a’b’ dark.
Now as usual rotating views find TL and
it’s inclinations.
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100
b
b1
LOCUS OF b & b1
Instead of considering a & a’ as projections of first point,
if v & VT’ are considered as first point , then true inclinations of line with
Hp & Vp i.e. angles  &  can be constructed with points VT’ & V respectively.
b’
b1’
a’
X
v
Y


VT’
a
b
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b
1
Then from point v & HT
angles  &  can be drawn.
&
From point VT’ & h’
angles  &  can be drawn.
PROBLEM :Line AB 100 mm long is 300 and 450 inclined to Hp & Vp respectively.
End A is 10 mm above Hp and it’s VT is 20 mm below Hp
.Draw projections of the line and it’s HT.
b1’
b’
FV
Locus of a & a1’
a’
10
SOLUTION STEPS:v
h’
X
Draw xy, one projector
0
(45 )
and locate on it VT and V.
20
 (300)
Draw locus of a’ 10 mm above xy.
VT’
Take 300 from VT and draw a line.
Where it intersects with locus of a’
HT
name it a1’ as it is TL of that part.
From a1’ cut 100 mm (TL) on it and locate point b1’
a
a1
0
Now from v take 45 and draw a line downwards
& Mark on it distance VT-a1’ I.e.TL of extension & name it a1
Extend this line by 100 mm and mark point b1.
TV
Draw it’s component on locus of VT’
& further rotate to get other end of Fv i.e.b’
Join it with VT’ and mark intersection point
(with locus of a1’ ) and name it a’
Now as usual locate points a and b and h’ and HT.
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a1’
Y
b
b1
PROBLEM :A line AB is 75 mm long. It’s Fv & Tv make
450 and 600 inclinations with X-Y line resp
End A is 15 mm above Hp and VT is 20 mm
below Xy line. Line is in first quadrant.
Draw projections, find inclinations with Hp &
Vp. Also locate HT.
FV
a’
Locus of a & a1’
15
X
b1’
b’
a1’
h’
v
600
20
450
Y


VT’
SOLUTION STEPS:HT
Similar to the previous only change
a
a1
is instead of line’s inclinations,
views inclinations are given.
So first take those angles from VT & v
TV
Properly, construct Fv & Tv of extension,
then determine it’s TL( V-a1)
and on it’s extension mark TL of line
and proceed and complete it. engineering108.com
b
b1
PROBLEM :- The projectors drawn from VT & end A of line AB are 40mm apart.
End A is 15mm above Hp and 25 mm in front of Vp. VT of line is 20 mm below Hp.
If line is 75mm long, draw it’s projections, find inclinations with HP & Vp
b’
a’
Draw two projectors for VT & end A
Locate these points and then
15
X
25
20
b1’
a1’
v
Y


VT’
a
40mm
b
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b1
GROUP (C)
CASES OF THE LINES IN A.V.P., A.I.P. & PROFILE PLANE.
Line AB is in AIP as shown in above figure no 1.
It’s FV (a’b’) is shown projected on Vp.(Looking in arrow direction)
Here one can clearly see that the
Inclination of AIP with HP = Inclination of FV with XY line
b’
a’



A.V.P.
A
B
Line AB is in AVP as shown in above figure no 2..
It’s TV (a b) is shown projected on Hp.(Looking in arrow direction)
Here one can clearly see that the
Inclination of AVP with VP = Inclination of TV with XY line
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a

b
LINE IN A PROFILE PLANE ( MEANS IN A PLANE PERPENDICULAR TO BOTH HP & VP)
For T.V.
ORTHOGRAPHIC PATTERN OF LINE IN PROFILE PLANE
VP
a’
PP
VT
a’
A
FV
b’
a”
LSV
b’
X
b”
HT
Y
a
B
TV
a
b
b
HP
Results:1. TV & FV both are vertical, hence arrive on one single projector.
2. It’s Side View shows True Length ( TL)
3. Sum of it’s inclinations with HP & VP equals to 900 (  +  = 900 )
4. It’s HT & VT arrive on same projector and can be easily located
From Side View.
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PROBLEM - Line AB 80 mm long, makes 300 angle with Hp
and lies in an Aux.Vertical Plane 450 inclined to Vp.
End A is 15 mm above Hp and VT is 10 mm below X-y line.
Draw projections, fine angle with Vp and Ht.
Locus of a’ & a1’
15
X
a’
450
VT
b1’
a1’
h’
v
10
b’ Locus of b’
Y


HT
a
Simply consider inclination of AVP
as inclination of TV of our line,
well then?
AVP 450 to VP
Locus of b’
b
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b1
PROBLEM :- A line AB, 75mm long, has one end A in Vp. Other end B is 15 mm above Hp
and 50 mm in front of Vp.Draw the projections of the line when sum of it’s
Inclinations with HP & Vp is 900, means it is lying in a profile plane.
Find true angles with ref.planes and it’s traces.
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(VT)
a”
Side View
( True Length )
VP
SOLUTION STEPS:After drawing xy line and one projector
VT
Locate top view of A I.e point a on xy as
a’
It is in Vp,
Front view
Locate Fv of B i.e.b’15 mm above xy as
it is above Hp.and Tv of B i.e. b, 50 mm
b’
below xy asit is 50 mm in front of Vp
a
Draw side view structure of Vp and Hp
X
and locate S.V. of point B i.e. b’’
From this point cut 75 mm distance on Vp and
Mark a’’ as A is in Vp. (This is also VT of line.) top view
From this point draw locus to left & get a’
Extend SV up to Hp. It will be HT. As it is a Tv
b
Rotate it and bring it on projector of b.
HT
Now as discussed earlier SV gives TL of line
and at the same time on extension up to Hp & Vp
gives inclinations with those panes.

HP
b”
(HT)
Y
APPLICATIONS OF PRINCIPLES OF PROJECTIONS OF LINES
IN SOLVING CASES OF DIFFERENT PRACTICAL SITUATIONS.
In these types of problems some situation in the field
or
some object will be described .
It’s relation with Ground ( HP )
And
a Wall or some vertical object ( VP ) will be given.
Indirectly information regarding Fv & Tv of some line or lines,
inclined to both reference Planes will be given
and
you are supposed to draw it’s projections
and
further to determine it’s true Length and it’s inclinations with ground.
Here various problems along with
actual pictures of those situations are given
for you to understand those clearly.
Now looking for views in given ARROW directions,
YOU are supposed to draw projections & find answers,
Off course you must visualize the situation properly.
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CHECK YOUR ANSWERS
WITH THE SOLUTIONS
GIVEN IN THE END.
ALL THE BEST !!
PROBLEM --Two objects, a flower (A) and an orange (B) are within a rectangular compound wall,
whose P & Q are walls meeting at 900. Flower A is 1M & 5.5 M from walls P & Q respectively.
Orange B is 4M & 1.5M from walls P & Q respectively. Drawing projection, find distance between them
If flower is 1.5 M and orange is 3.5 M above the ground. Consider suitable scale..
TV
B
A
FV
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Wall Q
PROBLEM :- Two mangos on a tree A & B are 1.5 m and 3.00 m above ground
and those are 1.2 m & 1.5 m from a 0.3 m thick wall but on opposite sides of it.
If the distance measured between them along the ground and parallel to wall is 2.6 m,
Then find real distance between them by drawing their projections.
TV
B
0.3M THICK
A
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PROBLEM -oa, ob & oc are three lines, 25mm, 45mm and 65mm
long respectively.All equally inclined and the shortest
is vertical.This fig. is TV of three rods OA, OB and OC
whose ends A,B & C are on ground and end O is 100mm
above ground. Draw their projections and find length of
each along with their angles with ground.
TV
O
C
A
FV
45 mm
B
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PROBLEM : A person observes two objects, A & B, on the ground, from a tower, 15 M high,
At the angles of depression 300 & 450. Object A is is due North-West direction of observer and
object B is due West direction. Draw projections of situation and find distance of objects from
observer and from tower also.
O
300
450
A
S
B
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W
PROBLEM :-Guy ropes of two poles fixed at 4.5m and 7.5 m above ground,
are attached to a corner of a building 15 M high, make 300 and 450 inclinations
with ground respectively.The poles are 10 M apart. Determine by drawing their
projections,Length of each rope and distance of poles from building.
TV
C
15 M
A
300
4.5 M
450
B
7.5M
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PROBLEM :- A tank of 4 M height is to be strengthened by four stay rods from each corner
by fixing their other ends to the flooring, at a point 1.2 M and 0.7 M from two adjacent walls respectively,
as shown. Determine graphically length and angle of each rod with flooring.
TV
4M
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PROBLEM :- A horizontal wooden platform 2 M long and 1.5 M wide is supported by four chains
from it’s corners and chains are attached to a hook 5 M above the center of the platform.
Draw projections of the objects and determine length of each chain along with it’s inclination with ground.
TV
Hook
H
D
A
C
B
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PROBLEM :A PICTURE FRAME 2 M WIDE AND 1 M TALL IS RESTING ON HORIZONTAL WALL RAILING
MAKES 350 INCLINATION WITH WALL. IT IS ATTAACHED TO A HOOK IN THE WALL BY TWO STRINGS.
THE HOOK IS 1.5 Ma ABOVE WALL RAILING. DETERMINE LENGTH OF EACH CHAIN AND TRUE ANGLE BETWEEN THEM
350
Wall railing
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Thank You
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